Problem of the Month (October 2003)

This month we are interested in power towers, stacks of powers of 2 in columns so that every column has the same digit sum. For example, here is a power tower with 4 columns, each with sum 10:

4096 51264 2 1

What is the fewest number of powers of 2 that we can arrange in n columns, each of which has sum s? What is the fewest number needed if 2k is one of the powers included? What if we use powers of 3, or powers of some other base?

Jeremy Galvagni and Bill Clagett considered the values of M(n,s), the the fewest number of powers of 2 that can be arranged in n columns with sum s. Here is Bill's data:

Values of M(n,s)

n \ s

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

1

1

1

2

1

2

2

3

1

2

2

3

2

3

3

4

2

3

3

4

3

2

2

2

2

2

4

2

4

2

4

4

4

3

5

4

5

4

6

5

5

5

3

3

3

4

3

4

3

4

3

5

5

5

4

6

5

5

5

8

7

8

7

4

4

4

4

4

5

3

5

4

6

5

6

5

7

6

5

5

5

6

5

6

5

4

5

6

6

6

5

6

6

6

6

6

7

5

5

6

7

7

7

8

7

8

5

6

8

8

8

8

8

8

6

9

9

9

10

9

10

10

10

10

11

11

11

12

12

12

13

13

Jeremy Galvagni conjectured that M(n,s) ≥ n, and that M(n,s) is increasing in n. Bill Clagett disproved both by finding this power tower:

65536
1024
2
1

Jeremy Galvagni also conjectured that M(n,8s)=ns, but then he found a counterexample M(3,16)=5. But he did manage to prove that M(an,s) ≤ a M(n,s) by noting that we can repeat the pattern, and possibily do better.